Ta thấy:\(\frac{x^7-x^4}{x^6-1}\)
\(=\frac{x^4\left(x^3-1\right)}{\left(x^2-1\right)\left(x^4+x^2+1\right)}\)
\(=\frac{x^4\left(x-1\right)\left(x^2+x+1\right)}{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x^4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x^4}{x^3+1}\)

\(=\frac{x^4-x^2-3x^2+3}{x^4-x^2+7x^2-7}=\frac{x^2\left(x^2-1\right)-3\left(x^2-1\right)}{x^2\left(x^2-1\right)+7\left(x^2-1\right)}=\frac{\left(x^2-3\right)\left(x^2-1\right)}{\left(x^2+7\right)\left(x^2-1\right)}=\frac{x^2-3}{x^2+7}\)

HELP ME

a) ĐKXĐ: 

\(\left\{{}\begin{matrix}x^2-9\ne0\\x+3\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne-3\end{matrix}\right.\Leftrightarrow x\ne\pm3\) 

b) \(A=\dfrac{x+15}{x^2-9}-\dfrac{2}{x+3}\)

\(A=\dfrac{x+15}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{x+15-2x+6}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{21-x}{\left(x+3\right)\left(x-3\right)}\)

c) Thay x = - 1 vào A ta có: 

\(A=\dfrac{21-\left(-1\right)}{\left(-1+3\right)\left(-1-3\right)}=\dfrac{21+1}{2\cdot-4}=\dfrac{22}{-8}=-\dfrac{11}{4}\)

Bài 10 trang 40 sgk toán lớp 8 tập ko 

\(\frac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\left(DK:x\ne-1;x\ne1\right)\)

\(=\frac{x^4\left(x^3+x^2+x+1\right)+\left(x^3+x^2+x+1\right)}{x^2-1}\)

\(=\frac{x^4\left[x\left(x^2+1\right)+x^2+1\right]+\left[x\left(x^2+1\right)+x^2+1\right]}{x^2-1}\)

\(=\frac{\left(x^4+1\right)\left(x+1\right)\left(x^2+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{\left(x^2+1\right)\left(x^4+1\right)}{x-1}\)

\(\frac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\)

\(=\frac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{\left(x+1\right)\left(x^6+x^4+x^2\right)}{\left(x+1\right)\left(x-1\right)}\)

\(=\frac{x^6+x^4+x^2}{x+1}\)

\(=\frac{x^2\left(x^3+x^2+1\right)}{x+1}\)

\(\dfrac{6x-x^2-5}{5x^6-x^7}\\ =\dfrac{-x^2+5x+x-5}{x^6\left(5-x\right)}\\ =\dfrac{\left(-x^2+5x\right)+\left(x-5\right)}{x^6\left(5-x\right)}\\ =\dfrac{-x\left(x-5\right)+\left(x-5\right)}{-x^6\left(x-5\right)}\\ =\dfrac{\left(-x+1\right)\left(x-5\right)}{-x^6\left(x-5\right)}\\ =\dfrac{-x+1}{-x^6}\)

\(\dfrac{-x^2+6x-5}{5x^6-x^7}\)

\(=\dfrac{x^2-6x+5}{x^7-5x^6}\)

\(=\dfrac{\left(x-1\right)\left(x-5\right)}{x^6\left(x-5\right)}\)

\(=\dfrac{x-1}{x^6}\)

Bài 1 : 

\(\left(x-2\right)^2-\left(x-3^2\right)=\left(x-2\right)^2-\left(x-9\right)\)

\(=x^2-4x+4-x+9=x^2-5x+13\)

Bài 2 : 

a, \(P=\frac{1-4x^2}{4x^2-4x+1}=\frac{\left(1-2x\right)\left(2x+1\right)}{\left(2x-1\right)^2}\)

\(=\frac{-\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)^2}=\frac{-\left(2x+1\right)}{2x-1}=\frac{-2x-1}{2x-1}\)

b, Thay x = -4 ta được : 

\(\frac{-2.\left(-4\right)-1}{2.\left(-4\right)-1}=\frac{8-1}{-8-1}=-\frac{7}{9}\)